Chapter 12.6, Problem 45HP

To determine

Calculate the volume of a sphere that is circumscribed about the cube.

Answer to Problem 45HP

The volume of the sphere is $586.6inc{h}^3$ .

Explanation of Solution

Given:

It is given in the question that a cube has a volume of $216inc{h}^3$ .

Concept Used:

In this, use the concept of Pythagoras theorem, volume of cube and volume of sphere i.e. $V_{cube}=a^3,V_{sphere}=\frac{4}{3}\pi r^3$ .

Calculation: Here, $V_{cube}=216inc{h}^3$

  $\begin{array}{l}{V}_{cube}=216inc{h}^3\\ a^3=216\\ a=\sqrt[3]{216}\\ a=6\end{array}$

  

Since a circumscribed sphere will only intersect the cube at its vertices. The center of the sphere is the center of the cube.

CQ = QR= RP = 3 because they are half of the side of the cube. Using Pythagoras theorem, to find CR.

  $\begin{array}{l}{3}^2+3^2=C{R}^2\\ C{R}^2=9+9\\ C{R}^2=18\\ CR=\sqrt{18}=3\sqrt{2}\end{array}$

Again using the Pythagoras theorem to find CP , the radius of the circle.

  $\begin{array}{l}$ {(3\sqrt{2})}^2+3^2=C{P}^2$C{P}^2=18+9\\ C{P}^2=27\\ CP=\sqrt{27}=3\sqrt{3}\end{array}$

Now, the volume of the sphere is

  $V_{sphere}=\frac{4}{3}\pi r^3=\frac{4}{3}\pi {(3\sqrt{3})}^3=\frac{4}{3}\pi (140.13)=\frac{1760.0}{3}=586.6inc{h}^3$

Conclusion:

The volume is $586.6inc{h}^3$ .